What combinations of vertical, horizontal, and oblique asymptotes are not possible for a rational function?

1 Answer
George C.
Dec 13, 2017

See explanation...

Explanation:

A rational function #y = (P(x))/(Q(x))#, where #P(x)# and #Q(x)# are non-zero polynomials, may have #0# or more vertical asymptotes, but the number of asymptotes must be finite.

It may also have a finite number of holes. This can happen when the numerator and denominator have common factors.

Example: #" "y = ((x-1)(x-2)(x-3))/((x-1)(x-2)(x-4)) = (x^3-6x^2+11x-6)/(x^3-7x^2+14x-8)#

In addition, any one of the following three possibilities must hold:

  • The function has no horizontal or oblique asymptotes.

Example: #" "y=x^2#

  • The function has a horizontal asymptote of the form #y=c# to which it tends for both #x->oo# and #x->-oo#. This can happen when the degree of the numerator is equal to that of the denominator.

Example: #" "y=(cx^2)/(x^2+1)#

  • The function has an oblique asymptote of the form #y=mx+c# to which it tends for both #x->oo# and #x->-oo#. This can happen when the degree of the numerator is exactly one greater than the degree of the denominator.

Example: #" "y = (mx^3+cx^2)/(x^2+1)#

So the following combinations are not possible:

#color(red)(xx)# The function has an infinite number of vertical asymptotes.

#color(red)(xx)# The function has an infinite number of holes.

#color(red)(xx)# The function has more than one horizontal asymptote.

#color(red)(xx)# The function has more than one oblique asymptote.

#color(red)(xx)# The function has a horizontal asymptote and an oblique asymptote.

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